Just a Hook

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 44248    Accepted Submission(s): 21158

 

Problem Description

In the game of DotA, Pudge’s meat hook is actually the most horrible thing for most of the heroes. The hook is made up of several consecutive metallic sticks which are of the same length.

Now Pudge wants to do some operations on the hook.
Let us number the consecutive metallic sticks of the hook from 1 to N. For each operation, Pudge can change the consecutive metallic sticks, numbered from X to Y, into cupreous sticks, silver sticks or golden sticks.
The total value of the hook is calculated as the sum of values of N metallic sticks. More precisely, the value for each kind of stick is calculated as follows:

For each cupreous stick, the value is 1.
For each silver stick, the value is 2.
For each golden stick, the value is 3.

Pudge wants to know the total value of the hook after performing the operations.
You may consider the original hook is made up of cupreous sticks.

Input

The input consists of several test cases. The first line of the input is the number of the cases. There are no more than 10 cases.
For each case, the first line contains an integer N, 1<=N<=100,000, which is the number of the sticks of Pudge’s meat hook and the second line contains an integer Q, 0<=Q<=100,000, which is the number of the operations.
Next Q lines, each line contains three integers X, Y, 1<=X<=Y<=N, Z, 1<=Z<=3, which defines an operation: change the sticks numbered from X to Y into the metal kind Z, where Z=1 represents the cupreous kind, Z=2 represents the silver kind and Z=3 represents the golden kind.

Output

For each case, print a number in a line representing the total value of the hook after the operations. Use the format in the example.

Sample Input

1
10
2
1 5 2
5 9 3

Sample Output

Case 1: The total value of the hook is 24.

 Source

2008 “Sunline Cup” National Invitational Contest

問題連結：HDU1698 Just a Hook

題解：線段樹基礎：區間修改+區間查詢

AC的C語言程式碼：

#include<stdio.h>
#include<string.h>
#define N 100005
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
int sum[N<<2],add[N<<2],a[N];

//更新函式
void PushUp(int rt)
{
	sum[rt]=sum[rt<<1]+sum[rt<<1|1];
} 

//建樹
void Build(int l,int r,int rt)
{
	if(l==r){
		sum[rt]=a[l];
		return;
	}
	int m=(l+r)>>1;
	Build(ls);
	Build(rs);
	PushUp(rt);//更新本結點 
}

//下推標記函式
void PushDown(int rt,int ln,int rn)//ln和rn分別表示rt的左右子樹的數量 
{
	if(add[rt]){
		//下推標記 
	    add[rt<<1]=add[rt<<1|1]=add[rt];
		//修改子結點的區間資訊sum，使之與對應的add相對應 
		sum[rt<<1]=add[rt]*ln;
		sum[rt<<1|1]=add[rt]*rn;
		//清除本結點的標記 
		add[rt]=0; 
	}
}
//區間修改
void Update(int L,int R,int C,int l,int r,int rt)
{
	if(L<=l&&r<=R){
		sum[rt]=C*(r-l+1);//更新數字和，向上保持正確 
		add[rt]=C;//增加標記，表示本區間的sum正確 
		return;
	}
	int m=(l+r)>>1;
	PushDown(rt,m-l+1,r-m);//下推標記 
	if(L<=m)
	  Update(L,R,C,ls);
	if(R>m)
	  Update(L,R,C,rs);
	PushUp(rt); 
} 

//區間查詢
int Query(int L,int R,int l,int r,int rt)
{
	if(L<=l&&r<=R)
	  return sum[rt];
	int m=(l+r)>>1;
	//下推標記，否則sum可能不正確
	PushDown(rt,m-l+1,r-m);
	int ans=0;
	if(L<=m)
	  ans+=Query(L,R,ls);
	if(R>m)
	  ans+=Query(L,R,rs);
	return ans;
}

int main()
{
	int n,q,i,j,x,y,z,t;
	scanf("%d",&t);
	for(i=1;i<=t;i++){
		memset(add,0,sizeof(add));//初始化標記陣列 
		scanf("%d%d",&n,&q);
		for(j=1;j<=n;j++)
		  a[j]=1;
		Build(1,n,1);
		while(q--){
			scanf("%d%d%d",&x,&y,&z);
			Update(x,y,z,1,n,1);
		}
		printf("Case %d: The total value of the hook is %d.\n",i,Query(1,n,1,n,1));	
	}
	return 0;
}